Category: Algebra
"Published in Newark, California, USA"
Find the factor:
(x + 2y)3 - (x3 + 8y3) = 0
Solution:
If you notice that the second group is a sum of two cubes. We can factor the second group as follows
(x + 2y)3 - (x3 + 8y3) = 0
(x + 2y)3 - (x + 2y)(x2 - 2xy + 4y2) = 0
Take out (x + 2y) as their common factor in the above equation
(x + 2y)[(x + 2y)2 - (x2 - 2xy + 4y2)] = 0
If you expand and simplify the second group, x2 and y2 will be cancel
(x + 2y)[x2 + 4xy +4y2 - x2 + 2xy - 4y2] = 0
(x + 2y)(6xy) = 0
Therefore, the answer is (x + 2y)(6xy) = 0
This website will show the principles of solving Math problems in Arithmetic, Algebra, Plane Geometry, Solid Geometry, Analytic Geometry, Trigonometry, Differential Calculus, Integral Calculus, Statistics, Differential Equations, Physics, Mechanics, Strength of Materials, and Chemical Engineering Math that we are using anywhere in everyday life. This website is also about the derivation of common formulas and equations. (Founded on September 28, 2012 in Newark, California, USA)
Saturday, December 1, 2012
Friday, November 30, 2012
Proving Two Parallel Lines - Circles
Category: Plane Geometry
"Published in Newark, California, USA"
If two circles are tangent externally and a line is drawn through a point of contact and terminated by the circles. Prove that the radii drawn to its extremities are parallel.
Solution:
Consider the given figure
Proof:
1. Statement: ∠1 ≅ ∠2
Reason: Vertical angles are congruent.
2. Statement: OP ≅ OA and O'P ≅ O'B
Reason: All points in a circle are equidistant from its center.
3. Statement: AB is drawn through point P.
Reason: Given item.
4. Statement: ΔOAP and ΔO'PB are isosceles triangles.
Reason: An inscribed triangle in a circle which consist of a center of a circle and the two end points of a chord is always an isosceles triangle.
5. Statement: ∠1 ≅ ∠3 and ∠2 ≅ ∠4
Reason: The two opposite angles of an isosceles triangle are congruent.
6. Statement: ∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4
Reason: Transitive property of congruence.
7. Statement: ∠AOP = 180º - (∠1 + ∠3)
∠PO'B = 180º - (∠2 + ∠4)
Reason: The sum of the interior angles of a triangle is 180º.
8. Statement: ∠AOP ≅ ∠PO'B
Reason: By computation at #7, if ∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4, then ∠AOP ≅ ∠PO'B.
9. Statement: OA ║ O'B
Reason: If a transveral line (OO') passed the two alternating interior angles (∠AOP and ∠PO'B) that are congruent, then it follows that the two lines (OA and O'B) which are adjacent to the alternating interior angles are parallel.
"Published in Newark, California, USA"
If two circles are tangent externally and a line is drawn through a point of contact and terminated by the circles. Prove that the radii drawn to its extremities are parallel.
 |
| Photo by Math Principles in Everyday Life |
Consider the given figure
|
| Photo by Math Principles in Everyday Life |
Proof:
1. Statement: ∠1 ≅ ∠2
Reason: Vertical angles are congruent.
2. Statement: OP ≅ OA and O'P ≅ O'B
Reason: All points in a circle are equidistant from its center.
3. Statement: AB is drawn through point P.
Reason: Given item.
4. Statement: ΔOAP and ΔO'PB are isosceles triangles.
Reason: An inscribed triangle in a circle which consist of a center of a circle and the two end points of a chord is always an isosceles triangle.
5. Statement: ∠1 ≅ ∠3 and ∠2 ≅ ∠4
Reason: The two opposite angles of an isosceles triangle are congruent.
6. Statement: ∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4
Reason: Transitive property of congruence.
7. Statement: ∠AOP = 180º - (∠1 + ∠3)
∠PO'B = 180º - (∠2 + ∠4)
Reason: The sum of the interior angles of a triangle is 180º.
8. Statement: ∠AOP ≅ ∠PO'B
Reason: By computation at #7, if ∠1 ≅ ∠2 ≅ ∠3 ≅ ∠4, then ∠AOP ≅ ∠PO'B.
9. Statement: OA ║ O'B
Reason: If a transveral line (OO') passed the two alternating interior angles (∠AOP and ∠PO'B) that are congruent, then it follows that the two lines (OA and O'B) which are adjacent to the alternating interior angles are parallel.
Thursday, November 29, 2012
Money - Investment Problem
Category: Algebra
"Published in Newark, California, USA"
Mr. Manalang has 3 children who are entering college. He was offered to pay 75% of the tuition fee for the second and 50% for the third if he enrolls his 3 children in the same school. If the total tuition fee amounts to ₱ 4,500.00, how much is the regular tuition fee?
Solution:
The above problem statement involves the money problem. Let's analyze the whole statements carefully as follows.
Let x be the amount of tuition fee for Mr. Manalang's 1st child.
Let 0.75 x be the amount of tuition fee for Mr. Manalang's 2nd child.
Let 0.50 x be the amount of tuition fee for Mr. Manalang's 3rd child.
If you add all the tuition fees of Mr. Manalang's children, the total amount is ₱ 4,500.00. The working equation for this problem will be
x + 0.75 x + 0.50 x = 4,500
2.25 x = 4,500
x = 2,000
Therefore, the regular tuition fee is ₱ 2,000.00. This is also the tuition fee for Mr. Manalang's 1st child.
Note: The monetary sign, ₱, is Philippine Pesos. The given word problem was in 1964.
"Published in Newark, California, USA"
Mr. Manalang has 3 children who are entering college. He was offered to pay 75% of the tuition fee for the second and 50% for the third if he enrolls his 3 children in the same school. If the total tuition fee amounts to ₱ 4,500.00, how much is the regular tuition fee?
Solution:
The above problem statement involves the money problem. Let's analyze the whole statements carefully as follows.
Let x be the amount of tuition fee for Mr. Manalang's 1st child.
Let 0.75 x be the amount of tuition fee for Mr. Manalang's 2nd child.
Let 0.50 x be the amount of tuition fee for Mr. Manalang's 3rd child.
If you add all the tuition fees of Mr. Manalang's children, the total amount is ₱ 4,500.00. The working equation for this problem will be
x + 0.75 x + 0.50 x = 4,500
2.25 x = 4,500
x = 2,000
Therefore, the regular tuition fee is ₱ 2,000.00. This is also the tuition fee for Mr. Manalang's 1st child.
Note: The monetary sign, ₱, is Philippine Pesos. The given word problem was in 1964.
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